Proof. We start by showing that $|\mathscr{C}_{t_0}(\rho)| < \infty$ almost surely. For this result, by Borel–Cantelli, it suffices to show that

(3.2) \begin{equation} \sum_{r=0}^{\infty}\mathbb{P}\big(\mathscr{C}_{t_0}(\rho)\cap V_r \not= \emptyset\big)<\infty.\end{equation}

We begin by denoting the set of non-backtracking paths of length r in the tree $(\mathcal{T}, \mathbb{F}(\mathcal{T\,\,}))$ started at the root as follows:

\begin{align*} C_r = \{v=(v_0, \dots, v_r)\,:\, v_i \in V_i, \ i \in \{0,\dots, r\} { \ \text{and}\ v_i\sim v_{i+1},\ i\in \{0,\dots,r-1\}}\}. \end{align*}

Note that, with this notation, $v_0=\rho$ and the number of paths of length r corresponds to the number of vertices in generation r, that is, $|C_r|=|V_r|$ .

Let $c=(v_0, \dots, v_r) \in C_r$ be a path of length r in $(\mathcal{T}, \mathbb{F}(\mathcal{T\,\,}))$ . For each $i\in \{0,\dots, r\}$ , by definition of $G_{t_0}$ , we have that the probability that the vertices $v_{i-1}$ and $v_i$ are connected in $G_{t_0}$ is

\begin{align*} 1- \exp\left(\!-\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_{i}} t_0\right)\!.\end{align*}

By monotonicity, we may assume that $r = 2k$ for some $k \in \mathbb{N}_0$ . Thus, conditioning on $(\mathcal{T}, \mathbb{F}(\mathcal{T\,\,}))$ , the probability that there is a vertex in generation r that is also in $\mathscr{C}_{t_0}(\rho)$ is

\begin{align*}\begin{split} \mathbb{P}_{\mathcal{T},\mathbb{F}}\big(\mathscr{C}_{t_0}(\rho)\cap V_r \not=\emptyset\big) &\leq \sum_{v \in C_r} \prod_{i=1}^{r}\big(1- \exp\big (\!-\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_{i}}t_0\big )\big) \\ & \leq \sum_{v \in C_r} \prod_{i=0}^{k-1}\big(1- \exp\big (\!-\lambda \mathcal{F}_{v_{2i}}\mathcal{F}_{v_{2i+1}}t_0\big )\big). \end{split}\end{align*}

Denote by $\mathbb{T}_{\leq r}$ and $\mathbb{F}_{\leq r}$ the $\sigma$ -algebras generated by the BGW tree and the fitness values up to generation r, respectively. Conditioning on $\mathbb{F}_{\le 2k-3}$ and $\mathbb{T}_{\le 2(k-1)}$ and using the independence structure of the weighted BGW,

\begin{align*}\begin{aligned} \mathbb{E}\big[ \mathbb{P}_{\mathcal{T},\mathbb{F}} & \big(\mathscr{C}_{t_0}(\rho)\cap V_r \not=\emptyset\big) \, |\, \mathbb{F}_{\le 2k -3},\mathbb{T}_{\le 2k-2}\big] \\ & \leq \sum_{v \in C_{2k-2}} \prod_{i=1}^{k-2}\big(1- \exp\big (\!-\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_{i}}t_0\big )\big) \mathbb{E}\Bigg [ \sum_{w \in C_2} \big (1- \exp \big (\!-\lambda \mathcal{F}_{w_0}\mathcal{F}_{w_1}t_0 \big ) \big ) \Bigg ] , \end{aligned}\end{align*}

where $w = (w_0,w_1,w_2)$ . Iterating this procedure gives

\begin{align*} \mathbb{P}\big(\mathscr{C}_{t_0}(\rho)\cap V_r \not=\emptyset\big) \leq \mathbb{E}\Bigg [ \sum_{w \in C_2} \big(1- \exp\big (\!-\lambda \mathcal{F}_{w_0}\mathcal{F}_{w_1}t_0\big )\big) \Bigg]^k . \end{align*}

As the summand in the last expectation is bounded by 1 and $\mathbb{E}[ |C_2|] \leq \mathbb{E}[ \xi]^2 < \infty$ , we can deduce by dominated convergence that we can choose $t_0$ small enough such that

\begin{align*} \mathbb{E}\Bigg[ \sum_{w \in C_2} \big(1- \exp\big(\!-\lambda \mathcal{F}_{w_0}\mathcal{F}_{w_1}t_0\big )\big) \Bigg] < 1. \end{align*}

Thus, $\mathbb{P}\big(\mathscr{C}_{t_0}(\rho)\cap V_{2k} \not=\emptyset\big)$ is summable in k. Together with the monotonicity of the process, this implies (3.2). It follows by the Borel–Cantelli lemma that, almost surely for r sufficiently large, $\mathscr{C}_{t_0}(\rho) \cap V_r = \emptyset$ with probability 1, so that $|\mathscr{C}_{t_0}(\rho)|$ is finite almost surely.

The same argument shows that almost surely for each $v \in V_\ell$ , we have that $\mathscr{C}_{t_0}(v)$ , restricted to the vertices in the subtree of $\mathcal{T}$ rooted at v, is finite. This immediately implies the statement of the lemma, as $\bigcup_{r = 0}^\ell V_r$ is finite almost surely.

Proof. Let $t_0$ be as given in Lemma 1. Since the initial configuration of the contact process is finite, we can find a sufficiently large and finite k such that $A \subset \bigcup_{r = 0}^k V_r$ , so that by the graphical construction we have that $X_t\subset \bigcup_{v\in\bigcup_{r=0}^k V_r} \mathscr{C}_{t_0}(v)$ for all $t\in [0,t_0] $ . Then, according to Lemma 1, we have that

\begin{align*} \mathbb{P} (|X_t|<\infty, \ \text{for all}\ t\in [0,t_0] )=1. \end{align*}

Let us now define a spanning subgraph $G_{2t_0}$ of $\mathcal{T}$ as follows: two neighbouring vertices u, v in $\mathcal{T}$ with $d(\rho,u) < d(\rho,v)$ are connected if

\begin{align*} N_{(u,v)} ([t_0,2t_0] )\geq 1. \end{align*}

As before, let $\mathscr{C}_{t_0}(\rho)$ be the connected component of $\rho$ in the graph $G_{t_0}$ . For each $v\in \mathscr{C}_{t_0}(\rho)$ , we denote by $\mathscr{C}_{2t_0}(v)$ the connected component of v in the graph $G_{2t_0}$ . We observe that by the graphical construction

\begin{align*} X_t\subset \bigcup_{v\in \mathscr{C}_{t_0}(\rho)}\mathscr{C}_{2t_0}(v), \ \text{for all}\ t\in [t_0,2t_0]. \end{align*}

Then we can argue that, by the independence of $G_{t_0}$ and $G_{2t_0}$ , and appealing again to Lemma 1, that the right-hand side is finite a.s., implying

\begin{align*} \mathbb{P} (|X_t|<\infty, \ \text{for all } t\in [t_0,2t_0] )=1. \end{align*}

Iterating the argument, we can conclude the proof.

Proof. We denote by $(\widetilde{X}_t) \sim \widetilde{\textbf{CP}}_{\rho^+}(\mathcal{T\,\,}_L^+;\, \mathbf{1}_{\rho})$ the following modification of the contact process $\textbf{CP}_{\rho^+}(\mathcal{T\,\,}_L^+;\, \mathbf{1}_{\rho})$ . The process $(\widetilde{X}_t)$ shares the same infection and recovery clocks as $(X_t)$ , except in the root $\rho$ . A recovery attempt at $\rho$ at time t is only valid if $\widetilde{X}_t = \mathbf{1}_{\rho}$ , namely, when there are no other infected vertices apart from $\rho$ and $\rho^+$ . Let $S_L$ and $\widetilde{S}_L$ be the first excursion time when the process $\textbf{CP}_{\rho^+}(\mathcal{T\,\,}_L^+;\, \mathbf{1}_{\rho})$ and the modified process $\widetilde{\textbf{CP}}_{\rho^+}(\mathcal{T\,\,}_L^+;\, \mathbf{1}_{\rho})$ reach state 0, respectively. An excursion to 0 of $(\widetilde{X}_t) \sim \widetilde{\textbf{CP}}_{\rho^+}(\mathcal{T\,\,}_L^+;\, \mathbf{1}_{\rho})$ started from the initial configuration $\widetilde{X}_0 = \mathbf{1}_{\rho}$ can be described as follows.

• • Possibility 1. We terminate if $\rho$ recovers before infecting any of its children. The probability that $\rho$ recovers before infecting any of its children conditionally on $(\mathcal{T}_L, \mathbb{F}(\mathcal{T}_L))$ is given by

  \begin{equation*} p\,:\!=\, \frac{1}{1 + \lambda \mathcal{F}_{\rho}\sum_{j=1}^{D}\mathcal{F}_{v_j}}. \end{equation*}
  Furthermore, the expected waiting time to see the recovery of $\rho$ is
  \begin{align*} \frac{1}{1 + \lambda \mathcal{F}_\rho \sum_{j=1}^{D}\mathcal{F}_{v_j}}=p,\end{align*}
  conditionally on the fitness and on the event that the recovery happens first.

• • Possibility 2. The root $\rho$ infects any of its children, say $v_i$ , before the recovery of $\rho$ . The vertex $v_i$ is selected with probability proportional to the fitness of the root’s children, i.e. with probability

  \begin{equation*} \frac{ \mathcal{F}_{v_i}}{\sum_{j=1}^{D}\mathcal{F}_{v_j}}. \end{equation*}
  Let $S_i$ be the first time that $\mathbf{CP}_\rho(\mathcal{T}_L,\mathbf{1}_{v_i})$ reaches the all-healthy state on $\bigcup_{i=1}^D \mathcal{T\,\,}_{L-1}^{v_i}$ . Then the expected waiting time until all subtrees of the root have recovered is $\mathbb{E}_{\mathcal{T}_L, \mathbb{F}}[S_i]$ , conditionally on vertex $v_i$ being infected first. When this excursion finishes, we come back to the initial state and follow either possibility 1 or possibility 2.

Note that we have reached state $\mathbf{0}$ as soon as the process follows possibility 1. Hence, by splitting according to the number of times the process follows possibility 2 before finally taking possibility 1, we obtain the expected excursion time to 0 of $(\widetilde{X}_t)$ , given the tree $({\mathcal{T}_L, \mathbb{F}(\mathcal{T}_L)})$ , as

\begin{align*} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}}\big [\,\widetilde{S}_L\big] &= \sum_{k=0}^{\infty} p (1-p )^k \bigg[(k+1) p + \frac{k }{ \sum_{i=1}^{D} \mathcal{F}_{v_i}} \sum_{j=1}^D \mathcal{F}_{v_j}\mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S_j ] \bigg]\\ &= \sum_{k=0}^{\infty} p (1-p )^k \bigg [(k+1) p + \frac{k D}{ \sum_{i=1}^{D} \mathcal{F}_{v_i}} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S ] \bigg ], \end{align*}

where in the last equality we use the notation

(4.1) \begin{equation} S= \frac{1}{D} \sum_{i=1}^D \mathcal{F}_{v_i} S_i, \end{equation}

where we assume that the different $S_i$ are coupled via the same graphical construction on the same underlying tree but have different initial conditions (but in fact only the marginal distributions matter, as we are only interested in the expectation of S). Calculating the series in the previous display explicitly, we obtain

(4.2) \begin{equation} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [\widetilde{S}_L ] = 1 + \lambda D \mathcal{F}_{\rho} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S ]. \end{equation}

Now we find a recursive upper bound for $\mathbb{E}_{\mathcal{T}_L, \mathbb{F}}[\widetilde{S}_L] $ by establishing a relationship between $ \mathbb{E}_{\mathcal{T}_L, \mathbb{F}}[S] $ and the stationary distribution of the contact process $\textbf{CP}_{\rho}(\mathcal{T}_L)$ . Denote by $S_{L-1}^{i}$ the first time when $\textbf{CP}_{\rho}(\mathcal{T\,\,}_{L-1}^{+, v_i};\, \mathbf{1}_{v_i})$ becomes 0 on $\mathcal{T\,\,}_{L-1}^{v_i}$ . For $\textbf{CP}_\rho(\mathcal{T\,\,}^{+,v_i}_{L-1})$ , the rate of leaving the state 0 and the expected return time to 0 are given by

\begin{equation*} q_i(\textbf{0}) \,:\!=\, \lambda \mathcal{F}_{\rho} \mathcal{F}_{v_i} \quad \text{and}\quad m_i({\textbf{0}}) \,:\!=\, (q_i(\textbf{0}) )^{-1} + \mathbb{E}_{\mathcal{T\,\,}_{L-1}^{+,v_i}, \mathbb{F}} \big[S_{L-1}^{i} \big]. \end{equation*}

Similarly, for the process $\textbf{CP}_{\rho}(\mathcal{T}_L)$ we also find these two quantities:

\begin{align*}\begin{split} q(\textbf{0}) \,:\!=\, \lambda \mathcal{F}_{\rho} \sum_{i=1}^{D} \mathcal{F}_{v_i}\quad \text{and}\quad m({\textbf{0}}) \,:\!=\, \frac{1}{q(\textbf{0})} + \sum_{i=1}^{D} \frac{\lambda \mathcal{F}_\rho \mathcal{F}_{v_i}}{q(\textbf{0})} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S_i ], \end{split} \end{align*}

where we recall that $S_i$ is the first time when $\textbf{CP}_{\rho}(\mathcal{T}_L;\, \mathbf{1}_{v_i})$ reaches the completely recovered state 0 on $\bigcup_{i=1}^D \mathcal{T\,\,}_{L-1}^{v_i}$ . Let $\widetilde{\nu}_{\mathcal{T}_L}$ and $\nu_{\mathcal{T}_{v_i}}$ be the stationary distribution of $\textbf{CP}_{\rho}(\mathcal{T}_L)$ and $\textbf{CP}_{\rho}(\mathcal{T\,\,}_{L-1}^{+,v_i}),$ respectively. Therefore, we have

\begin{equation*} \widetilde{\nu}_{\mathcal{T}_L}(\textbf{0}) \,=\, \frac{1}{q(\textbf{0})m(\textbf{0})}\,=\, \frac{1}{1+\lambda D \mathcal{F}_{\rho} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}}\left[S \right]}. \end{equation*}

Note that, also in $\textbf{CP}_{\rho}(\mathcal{T}_L)$ , the process restricted to each subtree $\mathcal{T\,\,}_{L-1}^{v_i}$ evolves independently, so that $\widetilde{\nu}_{\mathcal{T}_L}: = \otimes_{j=1}^D \nu_{\mathcal{T}_{v_j}}$ , which yields

(4.3) \begin{equation} \widetilde{\nu}_{\mathcal{T}_L}(\textbf{0})\,=\,\frac{1}{1+\lambda D \mathcal{F}_{\rho} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S ]} \,=\, \prod_{i=1}^{D}\frac{1}{1+\lambda \mathcal{F}_{\rho} \mathcal{F}_{v_i}\mathbb{E}_{\mathcal{T\,\,}_{L-1}^{+,v_i}, \mathbb{F}} [S^{i}_{L-1} ]}. \end{equation}

Using this fact, we obtain from (4.2) that

\begin{equation*} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [\widetilde{S}_L ] = 1 + \lambda D \mathcal{F}_{\rho} \mathbb{E}_{\mathcal{T}_L, \mathbb{F}} [S ] = \prod_{i=1}^{D} \Big (1+\lambda \mathcal{F}_{\rho} \mathcal{F}_{v_i}\mathbb{E}_{\mathcal{T\,\,}_{L-1}^{+,v_i}, \mathbb{F}} [S_{L-1}^{i} ] \Big ). \end{equation*}

Denote by $\widetilde S_{L-1}^{i}$ the first time when $\widetilde{\textbf{CP}}_{\rho}(\mathcal{T\,\,}^{+,v_i}_{L-1};\, \mathbf{1}_{v_i})$ reaches 0, where this process is defined as $\textbf{CP}_{\rho}(\mathcal{T\,\,}^{+,v_i}_{L-1};\, \mathbf{1}_{v_i})$ but $v_i$ is only allowed to recover if the subtree of all its children are all healthy. By using the monotonicity of the contact process, we have $S_{L-1}^{i} \leq \widetilde S_{L-1}^{i}$ . Now, taking expectations conditionally only on D and $\mathcal{F}_\rho$ and using the result that $\widetilde S_{L-1}^{i}$ does not depend on $\mathcal{F}_\rho$ by construction, we get

\begin{align*} \begin{split} \mathbb{E}\big[\widetilde{S}_L \mid D,\mathcal{F}_{\rho}\big ] &\leq \mathbb{E}\left[\prod_{i=1}^{D}\big(1+ \lambda \mathcal{F}_{\rho} \mathbb{E}\big[\mathcal{F}_{v_i}\widetilde S_{L-1}^{i}| \mathcal{T}_{v_i}, \mathbb{F}(\mathcal{T}_{v_i})\big]\big )\big|\big.D, \mathcal{F}_{\rho}\right] \\ &= \prod_{i=1}^{D}\big(1+ \lambda \mathcal{F}_{\rho} \mathbb{E}\big[\mathcal{F}_{v_i}\widetilde S_{L-1}^{i}\big]\big ). \end{split} \end{align*}

Moreover, we have

\begin{align*}\begin{split} \mathbb{E}\big[\widetilde{S}_L \mid D,\mathcal{F}_{\rho}\big ] = \prod_{i=1}^{D}\big (1+ \lambda \mathcal{F}_{\rho} \mathbb{E}\big[\mathcal{F}_{v_i }\widetilde S_{L-1}^i\big]\big ) = \big (1+\lambda \mathcal{F}_{\rho} \mathbb{E}\big [\mathcal{F}_{\rho}\widetilde S_{L-1} \big ]\big )^D. \end{split}\end{align*}

Thus, taking expectations over D and $\mathcal{F}_\rho$ , we obtain

(4.4) \begin{equation} \mathbb{E}\big[\mathcal{F}_\rho \widetilde S_L \big] \leq \mathbb{E}\big [\mathcal{F}_\rho \big (1+\lambda \mathcal{F}_{\rho} \mathbb{E}\big [\mathcal{F}_{\rho}\widetilde S_{L-1}\big ]\big )^D\big ]. \end{equation}

Now we apply an inductive argument over L to bound $ \mathbb{E} [\mathcal{F}_\rho \widetilde S_L ]$ . We assume that $\mathbb{E}[\mathcal{F}_{\rho}(1+c\mathcal{F}_\rho)^{\xi}]=M<\infty$ for some $c>0$ , and define the following constants:

\begin{align*} K= M\max\Big \{\frac{1}{c}, 1\Big \},\qquad \lambda_0 = \frac{1}{2K}.\end{align*}

For the base case $L=0$ , note that $\widetilde S_0$ is an exponential random variable with parameter 1 that is independent of $\mathcal{F}_\rho$ . Therefore,

\begin{align*} \mathbb{E}\big[\mathcal{F}_\rho \widetilde S_0\big] = \mathbb{E} [\mathcal{F} ] \mathbb{E} \big[\widetilde S_0\big] \leq M, \end{align*}

so that the base case holds. Next we assume $\mathbb{E}[\mathcal{F}_\rho \widetilde S_{L-1}]\leq M$ . Then we have, for any $\lambda \le \lambda_0$ , that

\begin{equation*} \frac{\lambda \mathbb{E}\big[\mathcal{F}_{\rho}\widetilde S_{L-1} \big ]}{c} \leq \frac{ \lambda M }{c} \leq \frac{M}{2 c K} = \frac{1}{2 c \max\{1/c,1\}} \leq 1. \end{equation*}

Hence, since $g(x)=\mathcal{F}_{\rho}(1+x\mathcal{F}_\rho)^{D}$ for $x>0$ is an increasing function, we have, for any L and all $\lambda\leq \lambda_0$ ,

(4.5) \begin{equation} \begin{split} \mathbb{E}\big[\mathcal{F}_\rho \widetilde S_L \big] &\leq \mathbb{E}\big[g\big(\lambda \mathbb{E}[\mathcal{F}_\rho \widetilde S_{L-1}]\big)\big]\leq \mathbb{E} [ g(c) ] = \mathbb{E}\big[\mathcal{F}_{\rho}(1+c\mathcal{F}_\rho)^{D}\big] = M, \end{split} \end{equation}

where the first inequality follows from (4.4). Moreover, taking into account the monotonicity of the contact process, we have $S_L \leq \widetilde{S}_L$ , which yields, for any L and all $\lambda\leq \lambda_0$ ,

(4.6) \begin{equation} \begin{split} \mathbb{E} [\mathcal{F}_\rho R_L ] \le \mathbb{E} [\mathcal{F}_\rho S_L ] \leq \mathbb{E} [\mathcal{F}_\rho \widetilde S_L ] \leq M, \end{split} \end{equation}

as required.

Proof of Theorem 1. We consider $\lambda_0$ as defined in Lemma 2. Let $\lambda\in (0,\lambda_0]$ . Let R denote the first time when $ {\textbf{CP}}(\mathcal{T}; \mathbf{1}_{\rho})$ reaches state 0. Then, by the monotone convergence theorem and Lemmas 2, we obtain

\begin{align*} \mathbb{E}[\mathcal{F}_{\rho}R] = \mathbb{E}\Big [\lim_{L\to \infty} \mathcal{F}_{\rho}R_L\Big ] = \lim_{L\to \infty} \mathbb{E}[\mathcal{F}_{\rho}R_L] \le M. \end{align*}

Now since $\mathcal{F}_{\rho}R$ is a non-negative random variable, we have that $\mathcal{F}_{\rho}R<\infty$ almost surely. Therefore,

\begin{align*}\mathbb{P}\big(X_t \not= \textbf{0}\ \text{for all}\ t \geq 0\big) = \mathbb{P}(R=\infty) = \mathbb{P}(\mathcal{F}_{\rho} R=\infty)=0,\end{align*}

where in the second equality we have used that $\mathcal{F}_\rho > 0$ almost surely. We conclude that, for all $\lambda \leq \lambda_0$ , the process $(X_t)\sim {\textbf{CP}}(\mathcal{T};\, \mathbf{1}_{\rho})$ dies out almost surely.

Proof. It should be emphasised here that the notation $ \mathbb{P}_{\mathbb{F}}(\!\cdot\!)$ corresponds to the conditional probability given the fitness and also that we start with $v_0$ initially infected. Let r be an arbitrary non-negative integer. First, we need to establish some appropriate notation. We define the sequence of times $(s_i)_{i\geq 0}$ by setting $s_0=0$ and, for $i\in \{1, \dots, r\}$ , defining

\begin{equation*} s_i\,:\!=\, \inf \{s\geq s_{i-1}: \ v_{i-1}\ \text{recovers or infects}\ v_i \mbox{ at time } s\}. \end{equation*}

Denote $T=\sum_{i=1}^{r}t_i$ where $t_i\,:\!=\,s_i - s_{i-1}$ . Also denote the events

\begin{align*} B_i\,:\!=\, \{ v_{i-1} \text{ infects } v_i \text{ before recovering}\} \quad \text{and} \quad B=\bigcap_{i=1}^rB_i.\end{align*}

We begin by noting that, from the definition of the event B and the definition of T, we can obtain the following lower bound:

(5.2) \begin{equation} \mathbb{P}_{\mathbb{F}}\Bigg(v_r \in \bigcup_{s\leq 2r} X_{s} \Bigg) \geq \mathbb{P}_{\mathbb{F}}\big(B\cap \{ T \leq 2r \}\big)= \mathbb{P}_{\mathbb{F}}\big( T\leq 2r\, | \, B\big)\mathbb{P}_{\mathbb{F}}(B). \end{equation}

Now we establish lower bounds for the two probabilities on the right-hand side of this equation. Conditioning on the fitness and also on the event $\{v_{i-1}\in X_{s_{i-1}}\}$ , we know that the probability that $v_{i-1}$ infects $v_i$ before recovering is given by

\begin{equation*} \mathbb{P}_{\{\mathcal{F}_{v_{i-1}}, \mathcal{F}_{v_i}\}} (B_i \ |\ v_{i-1}\in X_{s_{i-1}} )= \frac{\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}{1+\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}, \end{equation*}

where here we denote by $\mathbb{P}_{\{\mathcal{F}_{v_{i-1}}, \mathcal{F}_{v_i}\}}(\!\cdot\!)$ the conditional probability $\mathbb{P}(\!\cdot |\{\mathcal{F}_{v_{i-1}}, \mathcal{F}_{v_i}\} )$ . This implies that

(5.3) \begin{equation} \mathbb{P}_{\mathbb{F}}(B) = \prod_{i=1}^{r} \mathbb{P}_{\{\mathcal{F}_{v_{i-1}}, \mathcal{F}_{v_i}\}} (B_i \ |\ v_{i-1}\in X_{s_{i-1}} ) = \prod_{i=1}^{r} \frac{\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}{1+\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}. \end{equation}

Conversely, by an application of Markov’s inequality and by the definition of B and T, for any $\theta >0$ ,

\begin{align*} \begin{aligned} \mathbb{P}_\mathbb{F} (T\geq 2r\ | B ) = \mathbb{P}_\mathbb{F}\big( \mathrm{e}^{\theta T} \geq \mathrm{e}^{2\theta r}\ \big | B \big)\nonumber &\leq \mathrm{e}^{-2\theta r}\mathbb{E}_{ \mathbb{F}}\big[\mathrm{e}^{\theta T}\ \big | \ B\big] \\ &\leq \mathrm{e}^{-2\theta r}\prod_{i=1}^{r}\mathbb{E}_{\{\mathcal{F}_{v_{i-1}}, \mathcal{F}_{v_i}\}}\big[\mathrm{e}^{\theta t_i}\ \big|\big. \ B_i\cap\{v_{i-1}\in X_{s_{i-1}}\}\big]. \end{aligned} \end{align*}

By conditioning on the event $B_i\cap\{v_{i-1}\in X_{s_{i-1}}\}$ , we obtain that $t_i$ has an exponential distribution with parameter $(1+\lambda \mathcal{F}_{v_{i-1}} \mathcal{F}_{v_i})$ . Therefore, we can couple $t_i$ with a random variable $\tau_i$ with an exponential distribution with parameter 1 such that $t_i \leq \tau_i$ almost surely. Then, following the standard argument for a large deviation bound, we obtain that

\begin{align*} \mathbb{P}_\mathbb{F} (T\geq 2r\ | \ B ) \leq \mathrm{e}^{-2 r \theta^* + r \log \phi(\theta^*)}, \end{align*}

where $\phi(\theta^*) = \mathbb{E}[ e^{\theta^* \tau_i}]$ . Now, note that

\begin{align*} \lim\limits_{\theta^* \to 0}\frac{\log(\phi_i(\theta^*))}{\theta^*} = \mathbb{E}[ \tau_i] = 1. \end{align*}

Therefore, by choosing $\theta^* > 0$ small enough, we can deduce that there exists $\gamma > 0$ such that

(5.4) \begin{equation} \mathbb{P}_\mathbb{F} (T\geq 2r\ | \ B ) \leq \mathrm{e}^{-r \gamma}. \end{equation}

Substituting (5.3) and (5.4) back into (5.2), we now see that

\begin{equation*} \mathbb{P}_{\mathbb{F}}\Bigg(v_r \in \bigcup_{s\leq 2r} X_{s} \Bigg)\geq \big(1-\mathrm{e}^{-\gamma r}\big) \prod_{i=1}^{r} \frac{\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}{1+\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}. \end{equation*}

For the second part, we fix $f\geq \vartheta$ . Then, on the event $\{\mathcal{F}_{v_0} \geq f, \, \mathcal{F}_{v_r}\geq f\}$ and using that $\mathcal{F}_{v_i} \geq \vartheta$ , we obtain

\begin{equation*} \prod_{i=1}^{r} \frac{\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}}{1+\lambda \mathcal{F}_{v_{i-1}}\mathcal{F}_{v_i}} \geq \left(\frac{\lambda\vartheta^2}{1+\lambda\vartheta^2}\right)^{r-2}\left(\frac{\lambda f \vartheta}{1+\lambda f \vartheta}\right)^2, \end{equation*}

which yields the desired result.

Proof. Define $\mathrm{e}^\theta= (1+\lambda f\vartheta/2)^{-1}$ and assume that $Z_n\in (0,L)$ . After the same straightforward calculations as in [Reference Huang and Durrett11, Lemma 2.1] (but with infection rate $\lambda f$ ), we obtain with $D= L+\lambda f\vartheta (k-L) +1 \ge 0$ , for $y>0$ ,

\begin{equation*} \begin{aligned} \mathbb{E}[\mathrm{e}^{\theta Z_{n+1}} - \mathrm{e}^{\theta Z_n} \mid Z_n=y] & = \frac{\mathrm{e}^{\theta y}}{D} \bigg((\mathrm{e}^{-\theta}-1)L + (\mathrm{e}^\theta -1)\lambda f\vartheta (k-L) + \frac{\mathrm{e}^{-\theta}-1}{1+\lambda f\vartheta - \mathrm{e}^{-\theta}}\bigg) \\ & = \frac{\mathrm{e}^{\theta y}}{D} \bigg(\frac{\lambda f \vartheta}{2}L - \frac{\lambda f \vartheta}{2+\lambda f \vartheta }\lambda f\vartheta (k-L) + 1\bigg) , \end{aligned} \end{equation*}

where we used the definition of $\theta$ . It remains to show that the term in the bracket on the right-hand side is negative. We write, using the previously given definition of L,

\begin{align*} \begin{aligned} \frac{\lambda f \vartheta}{2}L & - \frac{\lambda f \vartheta}{2+\lambda f \vartheta }\lambda f\vartheta (k-L) + 1 \\ & = \lambda f \vartheta\Big( \frac{2+3\lambda f \vartheta}{2(2+\lambda f \vartheta)}L - \frac{\lambda f \vartheta}{2 + \lambda f \vartheta} k \Big) + 1 \\ & \leq \lambda f \vartheta\bigg( \frac{2+3\lambda f \vartheta}{2(2+\lambda f \vartheta)}\Big(\frac{\lambda f \vartheta k}{1+2\lambda f \vartheta}+1\Big) - \frac{\lambda f \vartheta}{2 + \lambda f \vartheta} k \bigg) + 1 \\ & = - \frac{(\lambda f \vartheta)^3}{2(2+ \lambda f \vartheta)(1+2\lambda f \vartheta)} k + \lambda f \vartheta \frac{2+3\lambda f \vartheta}{2(2+\lambda f \vartheta)} + 1 . \end{aligned} \end{align*}

Therefore, by calculating the asymptotics, we can deduce that the right-hand side is negative if either $k \geq 7$ and f is large or if k is large.

Proof. Let $\lambda>0$ , $k\geq 1$ , and $f\geq \vartheta$ , and define

\begin{align*}K= \left\lceil \frac{\lambda \vartheta {f} \kern0.5pt k^{1/3}}{1+2\lambda \vartheta {f}} \right\rceil\end{align*}

Recall the definition of the constant L given in (5.6) and note that $K\leq L$ . We begin by observing that

\begin{equation*} \mathbb{P}_{(0,1)} (T_K < T_{0,0} ) \geq \prod_{j=0}^{K-1}\frac{(k-j)\lambda f \vartheta}{1+(k-j)\lambda f \vartheta + j}, \end{equation*}

where the term in the product corresponds to the probability that $|\Lambda_t^0|$ jumps upwards K times before either the root or one of the leaves recovers. From the latter inequality and using [Reference Durrett8, Lemma 3.4.3], we have

\begin{align*}\begin{split} \mathbb{P}_{(0,1)}\big(T_K > T_{0,0}\big) &\leq \prod_{j=0}^{K-1}1 - \prod_{j=0}^{K-1}\frac{(k-j)\lambda f \vartheta}{1+(k-j)\lambda f \vartheta + j} \leq \sum_{j=0}^{K-1} \frac{1+j}{1+(k-j)\lambda f \vartheta+ j}\\ &\leq \sum_{j=0}^{K-1} \frac{1+j}{(k-j)\lambda f \vartheta} \leq \frac{K^2}{(k-K)\lambda f \vartheta}, \end{split}\end{align*}

where in the last inequality we have used that $\{(1+j)/(k-j), j=0, \dots, K-1\}$ is increasing in j.

Note that, as the function $x \mapsto {(x k^{1/3})} / {(1+2x)}$ is increasing in x, we can upper bound the function by its limit as $x \rightarrow \infty$ , so that

\begin{align*} K \leq \Big\lceil \frac{k^{1/3}}{2} \Big\rceil \leq \frac 12 k^{1/3} + 1\leq \frac{3}{2} k^{1/3} , \end{align*}

as $k \geq 1$ . Moreover, note that $\frac{3}{2} k^{1/3} \leq \frac{3}{4} k$ for all $k \geq 7$ . Using these two estimates, we have

\begin{align*}\begin{split} \mathbb{P}_{(0,1)} (T_K > T_{0,0} ) & \leq \frac{K^2}{(k-K)\lambda f { \vartheta}} \leq c \frac{1}{\lambda f \vartheta k^{1/3}} , \end{split}\end{align*}

for $c = 4( \frac{3}{2})^2 =9$ .

Now, we use the jump process $(Z_n)$ and Lemma 4. We assume that k or f is large, so that $(1+\lambda f \vartheta/2)^{-Z_n}$ is a supermartingale while $Z_n \in (0,L)$ . We denote by $T_L^Z$ and $R_0^Z$ the analogous hitting time of a level above L and the return time to a level below 0 for the process $(Z_n)$ , respectively. Note that, since $(Z_n)$ is obtained from $(Y_t)$ by observing the process only at its jump times, we see that $\{ T_L^{Y} > R_0^{Y} \} = \{ T_L^{Z} > R_0^{Z} \}$ . By an application of the optional stopping theorem with the bounded stopping time $\tau \wedge n$ , where $\tau= R_0^Z \wedge T_L^Z$ , we get

\begin{align*} \mathbb{E}_{K}\big[ (1+\lambda f \vartheta/2 )^{-Z_{\tau\wedge n}}\big] \leq \mathbb{E}_{K}\big[ (1+\lambda f \vartheta/2 )^{-Z_{0}}\big]. \end{align*}

By letting $n\to \infty$ , we deduce

\begin{align*} \mathbb{E}_{K}\big[ (1+\lambda f \vartheta/2 )^{-Z_{\tau}}\big] \leq (1+\lambda f/2 )^{-K},\end{align*}

which implies

\begin{equation*} \mathbb{P}_{K} \big (R_{0}^Z < T_{L}^Z \big ) + \big (1-\mathbb{P}_{K}\big (R_{0}^Z < T_{L}^Z \big)\big ) (1+\lambda f \vartheta / 2 )^{-L} \leq (1+\lambda f \vartheta / 2 )^{-K}. \end{equation*}

Discarding the second term on the left-hand side, it follows that

(5.9) \begin{equation} \mathbb{P}_{K}\big (R_{0}^Z < T_{L}^Z \big)\leq (1+\lambda f \vartheta / 2 )^{-K}. \end{equation}

To go back to the unmodified process, we use that $\{ T_{0,0} < T_L \} \subset \{ R_0^Z < T_L^Z \}$ , so that we can apply (5.9). Combined with the strong Markov property, we obtain

\begin{align*} \mathbb{P}_{(0,1)} (T_L > T_{0,0} ) &\leq \mathbb{P}_{(0,1)} (T_K > T_{0,0} ) + \mathbb{P}_{(0,1)} (T_L > T_{0,0} \, |\, T_K < T_{0,0} ) \mathbb{P}_{(0,1)} (T_K < T_{0,0} )\\ &\leq \mathbb{P}_{(0,1)} (T_K > T_{0,0} ) + \mathbb{P}_{(K,1)} (T_L > T_{0,0} )\\ &\leq \mathbb{P}_{(0,1)} (T_K > T_{0,0} ) + \mathbb{P}_{K} (R_0^Z < T_L^Z ). \end{align*}

Then, combining with the previous estimates, we have the first claim of the lemma, that is,

\begin{equation*} \mathbb{P}_{(0,1)} (T_L > T_{0,0} ) \leq { c}\frac{1}{\lambda f \vartheta k^{1/3}} + (1+ \lambda f \vartheta / 2 )^{-K}. \end{equation*}

The first claim of the lemma now follows, if we note that

\begin{align*} (1+\lambda f \vartheta / 2 )^{-K} \leq (1+\lambda { f} \vartheta /2)^{- \frac{\lambda f \vartheta}{1+ { 2}\lambda f \vartheta}k^{1/3} } \leq (1+\lambda { f} \vartheta /2)^{-\frac 13 k^{1/3}} + \mathrm{e}^{- \frac 38 (\lambda f \vartheta)^2 k^{1/3}}, \end{align*}

where we used that if $\lambda f \vartheta \geq 1$ , as $x \mapsto {x} / ({1+2x})$ is increasing,

\begin{align*}\frac{\lambda f \vartheta}{1+ { 2}\lambda f \vartheta}k^{1/3} \geq \frac 13 k^{1/3}. \end{align*}

Moreover, if $\lambda f \vartheta \leq 1$ , then using that $\log(1 + x) \geq \frac 34 x$ for $x \leq 1/2$ ,

\begin{align*} \frac{\lambda f \vartheta}{1+ { 2}\lambda f \vartheta}k^{1/3} \log(1+ \lambda f \vartheta /2 ) \geq \frac{1}{8}(\lambda f \vartheta)^2 k^{1/3} .\end{align*}

To deal with the second claim of the lemma, recall the random variable $\mathfrak{N}$ defined in (5.5) with $\mathbb{E}_{0} [\mathfrak{N}] = (\lambda f \vartheta)^{-1}$ . From the definition of the process, we can calculate the drift of $Y_t$ before time $T_L^Y$ as

\begin{equation*} \mu_{\lambda, f}= -L + \lambda f \vartheta (k-L) - \frac{1}{\lambda f \vartheta}, \end{equation*}

so that $(Y_t - \mu_{\lambda, f} t)$ is a martingale when stopped at $T_L^Y$ .

An application of the optimal stopping theorem with the bounded stopping time $T_L^Y \wedge t$ tells us that

\begin{equation*} \mathbb{E}_{0}\big[Y_{T_L^Y \wedge t}\big] - \mu_{\lambda, f} \mathbb{E}_{0}\big[T_{L}^Y \wedge t\big] = \mathbb{E}_{0}[Y_0] = 0. \end{equation*}

Furthermore, if we can show that $\mu_{\lambda,f} > 0$ , since $\mathbb{E}_{0}[Y_{{ T_L^Y} \wedge t}] \leq L$ , we get the upper bound

\begin{equation*} \begin{split} \mathbb{E}_{0}\big [T_{L}^Y \wedge t\big] &\leq \frac{L}{\mu_{\lambda, f}}. \end{split} \end{equation*}

Moreover, by taking $t \to \infty$ , the same bound holds for $\mathbb{E}_{0} [T_{L}^Y ]$ .

Thus, it remains to upper bound $L / \mu_{\lambda,f}$ . Write $x=\lambda f \vartheta$ and consider

\begin{equation*} \begin{split} \mu_{\lambda, f} &= xk -L(1+x) -\frac{1}{x}\ge xk-\Big(\frac{xk}{1+2x}+1\Big)(1+x)-\frac{1}{x}\\ &= \frac{x^2 k}{1+2x}-(1+x) - \frac{1}{x}. \end{split} \end{equation*}

Therefore,

(5.10) \begin{equation} \mu_{\lambda,f} \ge \frac{1}{2} \frac{x^2 k}{1+2x} \end{equation}

holds if

\begin{align*} \frac{x^2 k}{1+2x}\ge 2\bigg(1+x + \frac{1}{x}\bigg) \quad \iff\quad k\ge 2\bigg(2+ \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\bigg) . \end{align*}

This condition can be guaranteed by either taking $k \geq 5$ and then f large enough (so that $x = \lambda f \vartheta$ is large) so that the right-hand side is smaller than 5. Alternatively, the condition can be achieved by taking $k \geq 18 \max\{ \frac{1}{x},1\} = 18\max\{ \frac{1}{\lambda f \vartheta},1\}$ .

In either case, we can deduce from (5.10) that

\begin{align*} \frac{L}{\mu_{\lambda, f}} \le \frac{\frac{xk}{1+2x}+1}{\frac{1}{2}\frac{x^2 k}{1+2x}}= 2 \bigg(\frac{1}{x}+\frac{2}{xk}+\frac{1}{x^2k}\bigg) \leq 2 \bigg(\frac{3}{x} + \frac{1}{x^2k}\bigg) { \le 8 \max\{ 1, x^{-2}\} } , \end{align*}

which gives the result claimed.

Proof. Fix $\lambda>0$ . Recall the process $(Y_t)_{t\ge 0}$ given in (5.7) and its corresponding stopping times $T_L^Y$ and $R_0^Y$ . We begin by letting the stochastic process $(\Gamma_t)_{t \geq 0}$ denote the number of infected leaves, where we ignore times when the root is not infected. Let us now introduce the stopping time $T_L^\Gamma = \inf\{t > 0\,:\, \Gamma_t \geq L \}$ . Then, by definition, we have $T_L \geq T_L^\Gamma$ a.s. Moreover, in the event that $T_L < T_{0,0}$ , we have that $T_L^\Gamma \leq T_L^{\,Y}$ .

Let us denote $g(k)=f \kern0.5pt k^{2/3}$ and choose $\delta>0$ as

\begin{align*}{\delta} \,:\!=\, \min \{ 1, \lambda f \vartheta \} / 4 . \end{align*}

Observe that

(5.11) \begin{equation} \begin{split} \mathbb{P}_{(0,1)}\big(T_L> g(k)\big) &\leq \mathbb{P}_{(0,1)}\big(T_L^\Gamma \geq\delta g(k)\big) + \mathbb{P}_{(0,1)}\big(T_L>g(k), \ T_L^\Gamma <\delta g(k)\big). \end{split} \end{equation}

Our objective now is to find upper bounds for both probabilities on the right-hand side. First, we deal with the first of these probabilities. Note that

\begin{align*} \begin{aligned} \mathbb{P}_{(0,1)}\big(T_L^\Gamma \geq \delta g(k)\big) & \leq \mathbb{P}_{(0,1)} ( T_L > T_{0,0} ) + \mathbb{P}_{(0,1)}\big(T_L^\Gamma \geq \delta g(k) , T_L < T_{0,0}\big) \\ & \leq c_1(\lambda, f, \vartheta, k) + \mathbb{P}_0( T_L^Y \geq \delta g(k) ), \end{aligned} \end{align*}

where we used the first part of Lemma 5. By the second part of the same lemma and Markov’s inequality, we also have that

\begin{align*} \mathbb{P}_0( T_L^Y \geq\delta g(k)\big) \leq \, \frac{\mathbb{E}_0[T_L^Y]}{\delta g(k)} \leq {\frac{8 \max\{ 1, (\lambda f \vartheta)^{-2}\}}{\delta g(k)}}. \end{align*}

Combining, we obtain

\begin{align*} \mathbb{P}_{(0,1)}\big(T_L^\Gamma \geq\delta g(k)\big) \leq {c_1(\lambda, f, \vartheta, k) + \frac{8 \max\{ 1, (\lambda f \vartheta)^{-2}\}}{\delta g(k)}}. \end{align*}

To estimate the second term in (5.11), let $\ell_t$ denote the amount of time until t that the process $(X_t^0)$ spends in the states when the root is not infected, i.e.

\begin{align*}\ell_t=\big|\big\{s \leq t\,:\, X_s^0=(j,0),\ \text{for some} \ j\in \{0,1, \dots, k \}\big\}\big|,\end{align*}

where $|\cdot|$ denotes the Lebesgue measure. Note that, as $T_L= T_L^{\Gamma} + \ell_{T_L}$ , it follows that

(5.12) \begin{equation} \mathbb{P}_{(0,1)} (T_L>g(k), \, T_L^\Gamma <\delta g(k) )\leq \mathbb{P}_{(0,1)} (T_L^\Gamma<\delta g(k), \, \ell_{T_L}>(1-\delta) g(k) ). \end{equation}

Now, we recall the random variable $\mathfrak{N}$ defined in (5.5). Denote by $m_t$ the number of times that the process $(\Gamma_t)$ makes a downward jump with the same distribution as $\mathfrak{N}$ until time t. Note that, if the current state is (i, 0) with $i\geq 1$ , then the time until either the next jump down happens or the root becomes re-infected is given by an exponential random variable with distribution Exp $(i(1+\lambda f \vartheta))$ . Such a random variable is stochastically dominated by another random variable that has distribution Exp $(1+\lambda f \vartheta)$ . Hence, each time period when the root is healthy can be dominated by

\begin{align*}\sum_{i=1}^{\mathfrak{N}+1}E_i,\end{align*}

where, for each $i=1, \dots, \mathfrak{N}+1$ , the random variables $E_i$ have independent Exp $(1+\lambda f \vartheta)$ distributions. Thus, in the event $\{T_L^\Gamma <\delta g(k)\}$ , we deduce that

\begin{align*}\ell_{T_L} \leq \sum_{j=1}^{m_{T_L^\Gamma}}\sum_{i=1}^{\mathfrak{N}_j+1}E_i^{(j)} \leq\sum_{j=1}^{m_{\delta g(k)}}\sum_{i=1}^{\mathfrak{N}_j+1}E_i^{(j)},\end{align*}

where the random variables $E_i^{(j)}$ have Exp $(1+\lambda f \vartheta)$ distributions and $\mathfrak{N}_j$ are independent random variables with the same distribution as $\mathfrak{N}$ . Further, as $\mathfrak{N}_{\mathfrak{j}}$ has a shifted geometric distribution, straightforward computations show that, for each j,

\begin{align*}\bar T_j\,:\!=\,\sum_{i=1}^{\mathfrak{N}_j+1}E_i^{(j)}\end{align*}

has an exponential distribution with parameter $\lambda f \vartheta$ . Denote $a\,:\!=\,(1-\delta)(2\delta)^{-1}$ . It follows by a large deviation upper bound (or Chernoff bound) that

\begin{align*} \mathbb{P}_{(0,1)} \big(T_L^\Gamma <\delta g(k), \, \ell_{T_L}>(1-\delta)g(k), \, m_{\delta g(k)} \leq 2\delta g(k)\big) \\ \leq \mathbb{P}_{(0,1)} \Bigg(\sum_{j=1}^{\lfloor 2\delta g(k)\rfloor} \bar T_j > (1-\delta)g(k)\Bigg) \leq \mathrm{e}^{-\lfloor 2\delta g(k) \rfloor I(a)}, \end{align*}

using that $\mathbb{E}[ \bar T_j] = 1/{(\lambda f \vartheta)} \leq (1-\delta)/(2 \delta)$ by the definition of $\delta$ and where I(a) is the rate function of an Exp $(\lambda f \vartheta)$ -random variable, i.e.

\begin{equation*} I(a) \,:\!=\,\sup_{t\ge 0}\left(ta - \log \mathbb{E}\big[\mathrm{e}^{t\bar T_j}\big]\right) = a\lambda f \vartheta -1 -\log\! (a\lambda f \theta). \end{equation*}

Note that, if $\lambda f \vartheta \geq 1$ , then $\delta = 1/4$ and $a = 3/2$ . Moreover, in this case using that $x - 1 - \log x \geq x / 20$ for $x \geq 3/2$ , we have that

\begin{align*} I(a) = \frac 32 \lambda f \theta - 1 - \log \bigg(\frac 32 \lambda f \theta\bigg) \geq \frac{1}{20} \, \lambda f \theta. \end{align*}

Conversely, if $\lambda f \vartheta \leq 1$ , then $\delta = \lambda f \vartheta / 4$ and

\begin{align*} a \lambda f \vartheta \geq 2 - \frac{1}{2} \lambda f \vartheta \geq \frac{3}{2} , \end{align*}

so that, by the same bound as before, $I(a) \geq a\lambda f\vartheta/20 \ge 3/40$ . Thus,

\begin{align*} \mathrm{e}^{-\lfloor 2\delta g(k)\rfloor I(a)} \;\le\; \begin{cases} \mathrm{e}^{\lambda f \vartheta/20}\mathrm{e}^{-\lambda f\vartheta g(k)/40}, & \text{if } \lambda f \vartheta \ge 1, \\[4pt] \mathrm{e}^{3/40} \mathrm{e}^{-3\lambda f\vartheta g(k)/80}, & \text{if } \lambda f \vartheta \le 1 . \end{cases} \end{align*}

If $g(k) = f \kern0.5pt k^{2/3}\ge 4$ then $\mathrm{e}^{\lambda f \vartheta/20}\mathrm{e}^{-\lambda f\vartheta g(k)/40}\le \mathrm{e}^{-\lambda f\vartheta f \kern0.5pt k^{2/3}/80}$ . Combining both cases, and using that $\mathrm{e}^{3/30}\leq 2$ , we deduce that, for k or f sufficiently large,

\begin{equation*} \begin{split} \mathbb{P}_{(0,1)}&\big(T_L^\Gamma <\delta g(k), \, \ell_{T_L}>(1-\delta)g(k), \, m_{\delta g(k)} < 2\delta g(k)\big) \leq \mathrm{e}^{-\lfloor 2\delta g(k)\rfloor I(a)}\leq 2\mathrm{e}^{-\lambda f\vartheta g(k)/80}. \end{split} \end{equation*}

In addition, note that the random variable $m_{\delta g(k)}$ has a Poisson distribution with parameter $\delta g(k)$ . Similarly, as before, using a large deviation bound (or Chernoff bound) we have

\begin{equation*} \mathbb{P}_{(0,1)} \big(m_{\delta g(k)}> 2 \delta g(k)\big) \le \mathrm{e}^{-I(a)}, \end{equation*}

where $a=2\delta g(k)$ and

\begin{equation*} I(a)= \sup\nolimits_{t\ge 0} (ta - \delta g(k)(\mathrm{e}^t-1) ) = \delta g(k) (2\log2-1 ). \end{equation*}

Therefore, with $\widehat{c}_1=2\log2 -1$ and $\widehat{c}_2 = 1/80$ ,

\begin{equation*} \begin{split} \mathbb{P}_{(0,1)}&\big(T_L^\Gamma < \delta g(k), \, \ell_{T_L}^\Gamma> (1-\delta) g(k)\big) \\ & \leq \mathbb{P}(m_{\delta g(k)}>2\delta g(k))+\mathbb{P}_{(0,1)}\big(T_L^\Gamma<\delta g(k), \, \ell_{T_L}>(1-\delta)g(k), \, m_{\delta g(k)} < 2\delta g(k)\big) \\ & \leq \mathrm{e}^{-\widehat{c}_1\delta g(k)} + {2\,} \mathrm{e}^{-\widehat{c}_2\lambda f\vartheta g(k)}. \end{split} \end{equation*}

Substituting the estimates back into (5.11), we obtain

\begin{equation*} \begin{split} \mathbb{P}_{(0,1)}\big(T_L>g(k)\big) &\leq \mathbb{P}_{(0,1)}\big(T_L^\Gamma >\delta g(k)\big) + \mathbb{P}_{(0,1)}\big(T_L>g(k), \, T_L^\Gamma <\delta g(k)\big) \\ & \leq {c_1(\lambda, f, \vartheta, k) + \frac{8 \max\{ 1, (\lambda f \vartheta)^{-2}\}}{\delta g(k)} } + \mathrm{e}^{-\widehat{c}_1\delta g(k)} + {2\,}\mathrm{e}^{-\widehat{c}_2\lambda f\vartheta g(k)}. \end{split} \end{equation*}

This concludes the proof.

Proof. Fix $\lambda >0$ . Here we use the notation $\mathbb{P} ( \!\cdot\! )\,:\!=\, \mathbb{P} (\! \cdot \! | \ \mathbb{F})$ . We begin by noting that on the event $\{\mathcal{F}_\rho\geq f\}$ and by monotonicity we have

\begin{align*} \bigg\{\inf_{f \kern0.5pt k^{2/3} \leq t \leq S} |\Lambda_t| \leq \epsilon L\bigg\} \subset \bigg\{\inf_{f \kern0.5pt k^{2/3} \leq t \leq S} |\Lambda_t^0| \leq \epsilon L\bigg\}. \end{align*}

Then it is enough to prove the estimate for the process $(\Lambda_t^0)_{t\ge 0}$ . By the strong Markov property applied at $T_L$ on the event that $T_L < f \kern0.5pt k^{2/3}$ and using that at $T_L$ the root is necessarily infected, we note that

(5.17) \begin{align} \mathbb{P}_{(0,1)}&\bigg(\inf_{f \kern0.5pt k^{2/3} \leq t \leq S} |\Lambda_t^0| \leq \epsilon L \bigg) \nonumber\\ &= \mathbb{P}_{(0,1)}\bigg(\inf_{f \kern0.5pt k^{2/3} \leq t \leq S} |\Lambda_t^0| \leq \epsilon L, \, T_L \geq f \kern0.5pt k^{2/3} \bigg) \mathbb{P}_{(0,1)}\bigg(\inf_{f \kern0.5pt k^{2/3} \leq t \leq S} |\Lambda_t^0| \leq \epsilon L, \, T_L < f \kern0.5pt k^{2/3} \bigg) \nonumber\\ &\leq \mathbb{P}_{(0,1)}\big(T_L \geq f \kern0.5pt k^{2/3}\big) + \mathbb{P}_{(L,1)}\Big(\inf_{0 \leq t \leq S} |\Lambda_t^0| \leq \epsilon L\Big). \end{align}

Now, appealing to Lemma 7, we have for any $\epsilon \in (0,1/2)$ and for either f or k large enough,

\begin{equation*} \mathbb{P}_{(L,1)}\Big( \inf_{0\leq t\leq S}|\Lambda_t^0| \leq \epsilon L \Big) \leq (3 + \lambda f \vartheta) (1+\lambda f \vartheta /2)^{-\epsilon L}. \end{equation*}

Now, if $k\geq 7$ is fixed we can choose f sufficiently large so that $L \geq \frac{9}{10}\lceil \frac{k}{2} \rceil$ and $\lambda f \vartheta \geq 1$ . Therefore,

\begin{align*}(3 + \lambda f \vartheta) (1+\lambda f \vartheta /2)^{-\epsilon L} \leq (3 + \lambda f \vartheta) (1+\lambda f \vartheta /2)^{-{\frac{5}{16}} \frac{9}{10}4 } \leq \hat{C_1} (2 + \lambda f \vartheta)^{-1/8} .\end{align*}

for some absolute constant $\widehat{C}_1$ . For k large, we use that $x \mapsto {x} / ({1+x})$ is increasing, so that

\begin{align*} (3 + \lambda f \vartheta) (1+\lambda f \vartheta /2)^{-\epsilon L} \leq (3 + \lambda f \vartheta) (1 + \lambda \vartheta^2/2)^{ - {\frac{5}{16}} \frac{\lambda\vartheta^2}{1+2\lambda \vartheta^2} k }. \end{align*}

Substituting these two bounds back into (5.17) and using Lemma 6, we get the desired result.

Proof. Let $m \,:\!=\, \lfloor \tilde{S}(2r+1)^{-1} \rfloor$ and $m_0 \,:\!=\, \lceil f \kern0.5pt k^{2/3}(2r+1)^{-1} \rceil$ . Since either f or k is large enough, we can assume that $f \kern0.5pt k^{2/3}\le S/4$ and w.l.o.g. $m_0\le m$ . Here we use the notation $\mathbb{P} ( \!\cdot\! )\,:\!=\, \mathbb{P} (\! \cdot \! | \ \mathbb{F})$ and we add the subscript $\mathbb{P}_{(0,1)}$ when only the root is initially infected. Begin by noting that

\begin{align*} \mathbb{P}_{(0,1)} ({u_{r}}\notin X_{s}\ \text{for all}\ s \in[0,\tilde{S}] ) \leq \mathbb{P}_{(0,1)}(\mathcal{B}^c) + \mathbb{P}_{(0,1)} ( {u_{r}}\notin X_{s}\ \text{for all}\ s \in[0,m(2r+1)], \, \mathcal{B} ), \end{align*}

where

\begin{align*} \mathcal{B}\,:\!=\,\Big\{\inf_{f \kern0.5pt k^{2/3}\leq s\leq S}|\Lambda_s| \geq \epsilon L\Big\} \end{align*}

and $|\Lambda_t|$ is the number of infected leaves of $\rho$ at time t. From Lemma 8, we know, for f and k large enough, that

\begin{equation*} \mathbb{P}_{(0,1)}\left(\mathcal{B}^c\right) \leq R(f,k,\lambda), \end{equation*}

where $R(f,k,\lambda)$ is defined in (5.16). The proof is thus complete as soon as we can show that, for f and r sufficiently large,

(5.19) \begin{equation} \begin{split} \mathbb{P}_{(0,1)} ( {u_{r}}\notin X_{s}\ \text{for all}\ s \in[0,m(2r+1)],\ \mathcal{B} ) \leq (1- \widehat{C}_{\lambda,f}\widehat{\lambda}^r )^m. \end{split} \end{equation}

Define the sequence of times $t_0=0$ and $t_i=(2r+1)i$ for $i \in \{1,\dots, m\}$ . For fixed $i \in \{0,\dots, m-1\}$ , let us define $\tau_i \,:\!=\, \inf\{u\ge t_i\,:\, \rho \in X_u \} $ ,

\begin{align*}B_{\tau_i} \,:\!=\, \Big \{\inf_{s \in [t_i, \tau_i\wedge t_{i+1}]} |\Lambda_{s}|\geq \epsilon L \Big \}\quad \text{and}\quad A_i\,:\!=\,\{u_{r}\notin X_{s}\ \text{for all}\ s \in[t_{i},t_{i+1}]\}.\end{align*}

Thus, with this notation we have

\begin{equation*} \mathbb{P}_{(0,1)} ({u_{r}}\notin X_{s}\ \text{for all}\ s \in[0,m(2r+1)] ,\, \mathcal{B} ) \leq \mathbb{P}_{(0,1)}\Bigg (\bigcap_{i=m_0}^{m-1}A_i, \bigcap_{i=m_0}^{m-1}B_{\tau_i} \Bigg ). \end{equation*}

Since $\mathcal{B}_{\tau_i}\in \mathcal{F}_{\tau_i}$ , $\tau_i$ is $\mathcal{F}_{\tau_i}$ -measurable and $\mathcal{F}_{t_i}\subset \mathcal{F}_{\tau_i}$ , we deduce

\begin{align*} \mathbb{P} (u_{r}\in X_{s}\ \text{for some}\ s & \in[t_i,t_{i+1}] \ \big|\big. \ B_{\tau_i}, \ \mathcal{F}_{t_i} ) \\ & \ge \mathbb{P} (\{u_{r}\in X_{s}\ \text{for some}\ s \in[\tau_i,t_{i+1}]\} \cap \{\tau_i\leq t_{i}+1\} \ \big|\big. \ B_{\tau_i}, \ \mathcal{F}_{t_i} ) \\ & \geq \mathbb{E} \big( \mathbf{1}_{\{ \tau_i \leq t_i+1 \} } \mathbb{P} \big(u_{r}\in X_{s}\ \text{for some}\ s \in[\tau_i,\tau_i + 2r ] \ \big|\big. \ \mathcal{F}_{\tau_i}\big) \, \big|\big. \, B_{\tau_i} , \, \mathcal{F}_{t_i}\big). \end{align*}

Now note that, by an application of the strong Markov property and monotonicity, we obtain

\begin{align*} \mathbb{P}(u_{r}\in X_{s}\ \text{for some}\ s \in[\tau_i,\tau_{i} + 2r] \mid \mathcal{F}_{\tau_i}) & \ge \mathbb{P}(u_{r}\in X_{s}\ \text{for some}\ s \in[0,2r]\mid X_0 = \{ \rho \}) \\ & \geq \big(1-\mathrm{e}^{-\gamma }\big) C_{\lambda, f} \widehat{\lambda}^r, \end{align*}

where we used Lemma 3 together with the fact that $1-\mathrm{e}^{-\gamma} \leq 1-\mathrm{e}^{-\gamma r} $ holds for $r\geq 1$ in the last step. In addition, under $B_{\tau_i}$ , the probability that $\tau_i \leq t_i+1$ is bounded from below by the probability that one of the L exponential clocks with rate $\lambda \vartheta f$ attached to the leaves rings before time 1, i.e.

\begin{align*}\mathbb{P}(\tau_i\leq t_i+1\mid B_{\tau_i}, \, \mathcal{F}_{t_i})\ge 1-\mathrm{e}^{-\lambda \vartheta f L}.\end{align*}

Combining the estimates, we have that

\begin{align*} \mathbb{P} (u_{r}\notin X_{s}\ \text{for all}\ s \in[t_i,t_{i+1}] \ \big|\big. \ B_{\tau_i}, \, \mathcal{F}_{t_i} ) \leq 1- (1-\mathrm{e}^{-\lambda \vartheta fL})(1-\mathrm{e}^{-\gamma})C_{\lambda, f}\widehat{\lambda}^r. \end{align*}

Therefore, by the tower property,

\begin{align*}\begin{aligned} \mathbb{P}_{(0,1)}\Bigg( & \bigcap_{i=m_0}^{m-1}A_i \cap B_{\tau_i} \Bigg)\\ & = \mathbb{E}_{(0,1)} \Big( \mathbf{1}_{\bigcap_{i=m_0}^{m-2} A_{i}\cap B_{\tau_i}} \mathbb{P} (u_{r}\not\in X_{s}\ \text{for all}\ s \in[t_{m-1},t_{m}] , B_{\tau_{m-1}} \big|\big. \ \mathcal{F}_{t_{m-1}} )\Big ) \\ & \leq \mathbb{E}_{(0,1)}\Big( \mathbf{1}_{\bigcap_{i=m_0}^{m-2} A_{i}\cap B_{\tau_i}} \mathbb{P} (u_{r}\not\in X_{s}\ \text{for all}\ s \in[t_{m-1},t_{m}] \big|\big. \ B_{\tau_{m-1}} ,\mathcal{F}_{t_{m-1}} ) \Big). \end{aligned}\end{align*}

Using the previous bound and iterating,

\begin{align*} \mathbb{P}_{(0,1)} ({u_{r}}\notin X_{s}\ \text{for all}\ s \in[0,m(2r+1)] , \mathcal{B} ) & \leq \prod_{i=m_0}^{m-1} \big(1- (1-\mathrm{e}^{-\lambda \vartheta fL})(1-\mathrm{e}^{-\gamma})C_{\lambda, f}\widehat{\lambda}^r\big)\\ & \leq \big (1- \widehat C_{\lambda, f}\widehat{\lambda}^r\big )^{m-m_0}. \end{align*}

Thus, the desired result follows, since $m-m_0\ge S/4(2r+1)-1$ .

Proof. We assume from now on that $\mathbb{P}$ refers to the conditional probability measure, given $A_{f,k}^{\rho}$ . Denote by $\mathbb{T}_{\le r}$ and $\mathbb{F}_{\le r}$ the $\sigma$ -algebras generated by the BGW tree and the fitness values up to generation r, respectively. We also denote by c(v) the set of children of vertex $v\in V(\mathcal{T}\kern2.6pt)$ . We begin by noting that

\begin{align*}\begin{split} \mathbb{E}\big[Z_r \mid \mathbb{T}_{\le r+1},\, \mathbb{F}_{\le r}\big] &= \sum_{v\in \mathcal{G}_r} \mathbf{1}_{\{\mathcal{F}_v\ge f\} } \mathbb{E}\big[\mathbf{1}_{\{\xi_v^\vartheta = k\}} \mid \mathbb{T}_{\le r+1},\, \mathbb{F}_{\le r}\big] \\ &= \sum_{w\in \mathcal{G}_{r-1}} \sum_{v \in c(w)} \mathbf{1}_{\{\mathcal{F}_v\ge f\} } \mathbb{E}\big[\mathbf{1}_{\{\xi^\vartheta_v = k\}} \mid \xi_v \big]. \end{split}\end{align*}

Now, using the tower property yields

\begin{align*}\begin{split} \mathbb{E} [Z_r \mid \mathbb{T}_{\le r},\, \mathbb{F}_{\le r-1} ] &= \mathbb{E}\big[\mathbb{E}\big[Z_r \mid \mathbb{T}_{\le r+1},\, \mathbb{F}_{\le r}\big]\mid \mathbb{T}_{\le r},\, \mathbb{F}_{\le r-1} \big] \\ &= \sum_{w\in \mathcal{G}_{r-1}} \sum_{v \in c(w)} \mathbb{E}\big[ \mathbf{1}_{\{\mathcal{F}_v\ge f\} } \mathbb{E}[ \mathbf{1}_{\{\xi^\vartheta_v = k\}} \mid \xi_v] \mid \mathbb{T}_{\le r},\, \mathbb{F}_{\le r-1} ] \big] \\ &= \sum_{w\in \mathcal{G}_{r-1}} \xi_w \mathbb{P}(\xi^\vartheta= k, \, \mathcal{F} \ge f). \end{split}\end{align*}

Thus, once again appealing to the tower property, we get

\begin{align*}\begin{split} \mathbb{E} [Z_r \mid \mathbb{T}_{\le r-1},\, \mathbb{F}_{\le r-2} ] &= \sum_{u\in \mathcal{G}_{r-2}} \sum_{w \in c(u)} \mathbb{E}[\xi \mathbf{1}_{\{\mathcal{F}\ge \vartheta\}}]\mathbb{P}(\xi^\vartheta= k, \, \mathcal{F} \ge f)\\ &= \sum_{u\in \mathcal{G}_{r-2}} \xi_u\mathbb{E} [\xi \mathbf{1}_{\{\mathcal{F}\ge \vartheta\}} ]\mathbb{P}( \xi^\vartheta= k, \, \mathcal{F} \ge f). \end{split}\end{align*}

Recursively, we deduce

\begin{align*}\begin{split} \mathbb{E} [Z_r \mid \mathbb{T}_{\le 2},\, \mathbb{F}_{\le 1} ] &= \sum_{u\in \mathcal{G}_{1}} \xi_u (\mathbb{E} [\xi \mathbf{1}_{\{\mathcal{F}\ge \vartheta\}} ] )^{r-2}\mathbb{P}(\xi^\vartheta= k, \, \mathcal{F} \ge f). \end{split}\end{align*}

Therefore,

\begin{align*}\mathbb{E}[Z_r]= \mu_\vartheta^{r-2}\mathbb{P}(\xi^\vartheta= k, \, \mathcal{F} \ge f) \mathbb{E}\Big[\sum_{u\in \mathcal{G}_{1}} \xi_u \Big] \ge k \mu_\vartheta^{r-2}\mathbb{P}(\xi^\vartheta= k, \, \mathcal{F} \ge f),\end{align*}

where in the last inequality we have used that we are working on the probability measure conditional on the event $A_{f,k}^{\rho}$ , which states that the set of children of $\rho$ with $\xi_v\ge 1$ and $\mathcal{F}_v\ge \vartheta$ has cardinality k.

Proof. Throughout this proof, fix $\lambda > 0$ . From now on, we assume that $\mathbb{P}$ refers to the conditional probability measure, given $A_{f,k}^{\rho}$ . Also, it must be emphasised that we always start with the root initially infected (unless specified otherwise). We first prove that, conditionally on $v\in V_r$ being a star, then it will be infected before time S with probability bounded away from zero uniformly for either f or k large enough. That is to say, there exists a positive constant $c_1$ such that for either f or k sufficiently large and any $v \in {\mathcal{A}_{f,k}^r}$ ,

\begin{equation*} p_\mathrm{in}\,:\!=\,\mathbb{P}_{\mathcal{T}, \mathbb{F}} (v \in X_s \ \text{for some}\ s\in [0,S] )\geq c_1. \end{equation*}

The proof of this bound follows by an application of Lemma 9, by using the monotonicity of the inhomogeneous contact process restricted to the subgraph consisting of a star with root $\rho$ together with a path $\mathcal{C}_r$ from one descendant of the root to v. Indeed, for either f or k sufficiently large, we obtain

\begin{equation*} 1-p_\mathrm{in}\le (1- \widehat{C}_{\lambda,f}\widehat{\lambda}^r )^{S/4(2r+1)-1} + R(f,k,\lambda), \end{equation*}

where the function $R(f,k,\lambda)$ is defined in (5.16). Then, using the inequality $(1-x)^{1/x}< \mathrm{e}^{-1}$ , observe that (6.5) forces the first term on the right-hand side of the last equation to be at most $\mathrm{e}^{-2}$ , i.e.

\begin{equation*} \big(1- \widehat{C}_{\lambda,f}\widehat{\lambda}^r\big)^{S/4(2r+1)-1} \leq \big(1- \widehat{C}_{\lambda,f}\widehat{\lambda}^r\big)^{2/(\widehat{C}_{\lambda, f} \widehat{\lambda}^r)}< \mathrm{e}^{-2}. \end{equation*}

Further, for either f or k sufficiently large, we have $R(f,k,\lambda)\leq 1/2$ . Therefore, we deduce that $p_\mathrm{in}$ is bounded away from 0 for either f or k large enough, i.e.

(6.7) \begin{equation} p_\mathrm{in} \geq c_1,\quad \text{where} \ c_1=\frac{1}{2}-\mathrm{e}^{-2}>0. \end{equation}

In other words, with positive probability, we push the infection to a generation r that satisfies Condition (6.5).

Proof. Let $k\ge 7$ and $f\ge \vartheta$ . First, we assume that ( B ) does not hold, i.e.

(6.13) \begin{equation} \mathbb{E}\big[(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{{\xi\ge 7}, \, \mathcal{F}\ge \vartheta\}}\big] < \infty, \,\, \text{for some}\,\, c>0. \end{equation}

Now, we note that

\begin{align*}\begin{split} \mathbb{E} [(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{ {\xi\ge 7}, \, \mathcal{F}\ge \vartheta\}} ] \ge \mathbb{E} [(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{\xi=k, \, \mathcal{F}\ge f\}} ] \ge (1+f)^{ck} \mathbb{P}(\xi=k,\, \mathcal{F} \geq f), \end{split}\end{align*}

which implies

\begin{align*}\begin{split} \limsup_{\substack{f+k\to \infty\\ {k\ge 7}, \, f\ge \vartheta}} \frac{\log (\mathbb{P}(\xi=k, \, \mathcal{F} \geq f) )}{k\log\! (1+f)} \leq \limsup_{\substack{f+k\to \infty\\ {k\ge 7}, \, f\ge \vartheta}} \frac{\log( (1+f)^{-ck} \mathbb{E}[(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{ {\xi\ge 7}, \, \mathcal{F}\ge \vartheta\}}])}{k \log\! (1+f)}=-c. \end{split}\end{align*}

In other words, we have shown that

(6.14) \begin{equation} \limsup_{\substack{f+k\to \infty\\ \, {k\ge 7}, \, f\ge \vartheta}} \frac{\log (\mathbb{P}(\xi=k, \, \mathcal{F} \geq f) )}{k\log\! (1 +f)}= -\delta, \,\, \text{for some} \ \delta\in (0,\infty]. \end{equation}

Now, assume that (6.14) holds. Then there exist $\delta>0$ and $C>0$ such that

(6.15) \begin{equation} \mathbb{P}(\xi=k, \, \mathcal{F} \geq f) \leq C(1+f)^{-\delta k},\ \text{for all} \ f\ge \vartheta \ \text{and}\ {k\ge 7}. \end{equation}

We choose $c\in (0,\delta)$ small enough, such that $2^c(1+\vartheta)^{c-\delta}<1$ . Now for some $k_0$ such that $(\delta-c)k_0>1$ , we observe that the following inequalities hold:

\begin{align*} \mathbb{E} [(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{\xi\geq k_0, \, \mathcal{F}\geq \vartheta\}} ] &\leq \sum_{k=k_0}^{\infty} \sum_{f= 0}^\infty \mathbb{E} [(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{\mathcal{F}\in [f+\vartheta, f+\vartheta+1)\}\cap \{\xi= k\}} ] \\ & \leq \sum_{k=k_0}^{\infty} \sum_{f= 0}^\infty (2+\vartheta+f)^{ck}\mathbb{P}(\xi =k, \, \mathcal{F}\geq f+\vartheta)\\ & \leq C\sum_{k=k_0}^{\infty} \sum_{f= 0}^\infty (2^c(1+\vartheta+f)^{-(\delta-c)} )^k \\ & = C\sum_{f= 0}^\infty \frac{2^{ck_0}(1+\vartheta+f)^{-(\delta-c) k_0}}{1-2^c(1+\vartheta+f)^{-(\delta-c)}}\\ & \le \frac{C\, 2^{ck_0}}{1-2^c(1+\vartheta)^{c-\delta}}\sum_{f= 0}^\infty (1+\vartheta+f)^{- (\delta-c) k_0} < \infty, \end{align*}

where in the third inequality we used that $(2+\vartheta+f)^{ck}\le 2^{ck} (1+\vartheta+f)^{ck}$ and in the equality we get a geometric sum that converges, since $2^c(1+\vartheta+f)^{c-\delta}\le 2^c(1+\vartheta)^{c-\delta}$ for all $f\ge 0$ . Conversely,

\begin{align*} \mathbb{E} [(1+\mathcal{F}\,)^{c\xi}\mathbf{1}_{\{{7\le \xi}\leq k_0, \, \mathcal{F}\geq \vartheta\}} ] &\leq \sum_{k=7}^{k_0} \int_0^\infty \mathbb{P}((1+\mathcal{F}\,)^{ck}\ge x, \, \xi =k, \, \mathcal{F}\ge \vartheta) \,\mathrm{d} x \\ & = \sum_{k=7}^{k_0}\int_{0}^{(1+\vartheta)^{ck}} \mathbb{P}(\xi =k, \, \mathcal{F}\ge \vartheta) \,\mathrm{d}x \\ & \quad + \sum_{k=7}^{k_0}\int_{(1+\vartheta)^{ck}}^\infty \mathbb{P}(\xi =k, \, \mathcal{F}\ge x^{1/ck}-1) \, \mathrm{d}x \\ & \le k_0 (1+\vartheta)^{ck_0} + C\sum_{k=7}^{k_0}\int_{1}^\infty x^{-\delta/c} \,\mathrm{d}x <\infty, \end{align*}

where in the last inequality we have used (6.15). Thus we get that (6.13) holds.

For the second claim, first suppose that the $\limsup$ in (6.12) is achieved along a sequence $(f_n,k_n)$ such that $f_n \rightarrow \infty$ . Then, we see that

\begin{align*}\begin{split} \mathbb{P}(\xi^\vartheta = k_n,\ & \mathcal{F}\ge f_n) = \sum_{\ell=1}^{\infty} \mathbb{P}\Bigg( \sum_{i=1}^{\ell} I_i = k_n \Bigg) \mathbb{P}(\xi =\ell, \, \mathcal{F}\ge f)\\ &\geq \mathbb{P}\Bigg(\sum_{i=1}^{k_n} I_i = k_n\Bigg) \mathbb{P}(\xi= k_n, \, \mathcal{F}\ge f_n) = \mathbb{P}(\mathcal{F}\geq \vartheta)^{k_n} \mathbb{P}(\xi =k_n, \, \mathcal{F}\ge f_n). \end{split}\end{align*}

Hence,

\begin{align*}\begin{split} \limsup_{\substack{f+k\to \infty\\ \, {k\ge 7}, \, f\ge \vartheta}} \frac{\log \big(\mathbb{P}(\xi^{\vartheta}=k, \, \mathcal{F} \geq f)\big)}{k\log\! (1+f)} \geq \lim_{n \rightarrow \infty} \frac{\log \big(\mathbb{P}(\mathcal{F}\geq \vartheta)^{k_n}\mathbb{P}(\xi=k_n, \, \mathcal{F}\geq f_n)\big)}{k_n\log\! (1+f_n)}=0, \end{split} \end{align*}

where in the last equality we have that $f_n \rightarrow \infty$ .

Now, suppose that the $\limsup$ in (6.12) is achieved along a sequence $(f_n,k_n)$ such that $k_n \rightarrow \infty$ . Set $p = \mathbb{P}(\mathcal{F} \geq \vartheta)$ . Then, we define $k_n^* \in \mathbb{N}$ such that $pk_n \leq k_n^* < p k_n + 1$ , so that $\lfloor k_n^*/p \rfloor = k_n$ . Similarly to before, we estimate

\begin{align*}\begin{split} \mathbb{P}(\xi^\vartheta = k_n^*, \, \mathcal{F}\ge f_n) &\geq \mathbb{P}\Bigg(\sum_{i=1}^{k_n} I_i = k_n^*\Bigg) \mathbb{P}(\xi= k_n, \, \mathcal{F}\ge f_n) . \end{split}\end{align*}

Now, since $I_i$ are independent Bernoulli (p) random variables, by the de Moivre–Laplace theorem, we have that, for $k_n$ large, there exists $c > 0$ such that

\begin{align*} \mathbb{P}\Bigg(\sum_{i=1}^{k_n} I_i = k_n^*\Bigg) = \frac{(1+o(1))}{\sqrt{p(1-p)k_n}} \exp\big\{\!- (k_n^* - p k_n)^2 / \sqrt{k_np(1-p)}\big\} \geq \frac{c}{\sqrt{k_n}} . \end{align*}

Thus, combining these estimates, we get that

\begin{align*} \frac{\log \big(\mathbb{P}(\xi^{\vartheta}=k_n^*, \, \mathcal{F} \geq f_n)\big)}{k_n\log\! (1+f_n)} \geq \frac{\log( c/\sqrt{k_n})}{k_n \log\! (1+f_n)} + \frac{\log \big(\mathbb{P}(\xi=k_n, \, \mathcal{F} \geq f_n)\big)}{k_n\log\! (1+f_n)} \rightarrow 0, \end{align*}

as $n\to \infty$ , which completes the proof.

Proof of Theorem 2. Throughout this proof, fix $\lambda > 0$ . We assume that $\mathbb{P}$ refers to the conditional probability measure, given $A_{f,k}^{\rho}$ , and as the latter event has positive probability it suffices to show that $\liminf_{t \rightarrow \infty} \mathbb{P}(\rho \in \xi_t)> 0$ under the conditional probability measure.

We would like to use Lemma 12 applied to the non-negative and non-decreasing function G defined in (6.6). First, observe that

\begin{equation*} \inf_{0 \leq s \leq S} g(s) \geq \mathbb{P} (\rho \ \text{never recovers in} \ [0,S] ) = \mathrm{e}^{-S} >0. \end{equation*}

Thus, we have that g satisfies the first condition of Lemma 12. Now we want to apply Lemma 13 with the following choice of r:

(6.16) \begin{equation} r ={r(f,k)=} \Bigg\lceil -\frac{\log\big(\mu^{-2}_{\vartheta} c k\mathbb{P}(\xi^\vartheta =k, \, \mathcal{F}\geq f)\big) }{\log \mu_\vartheta} \Bigg\rceil. \end{equation}

We can see that, as $\mathbb{E}[\xi^\vartheta]<\infty$ , we have by dominated convergence that

\begin{align*} k \mathbb{P}(\xi^\vartheta = k, \, \mathcal{F} \geq f) \leq \mathbb{E}[ \xi^\vartheta \mathbf{1}_{ \{ \xi^\vartheta \geq k , \mathcal{F} \geq f \} } ] \rightarrow 0, \end{align*}

as either $k \to \infty$ or $f \to \infty$ . Therefore, in either case $r = r(f,k)\rightarrow \infty$ . Moreover, we have $c k \mu^{r-2}_{\vartheta} \mathbb{P}\left(\xi^\vartheta =k, \, \mathcal{F}\geq f\right)\geq 1$ , and this in turn implies that $G(x)\geq x$ for some neighbourhood of 0. We complete our argument by showing that, under the hypothesis of the theorem, we can find f and k such that (6.5) holds. This allows us to apply Lemma 13 together with Lemma 12 to obtain

\begin{equation*} \liminf_{t \to \infty} \mathbb{P}(\rho \in X_t) >0 , \end{equation*}

first when conditioning on $A_{f,k}^\rho$ , but, since this event has positive probability, also without conditioning. Now by the reverse Fatou lemma, we conclude that

\begin{align*} \mathbb{P} ( \mbox{for all } s \geq 0 \mbox{ there exists }t \geq s \, : \, \rho \in X_{t} ) = \mathbb{E}\bigg[ \limsup_{t \rightarrow \infty} \mathbf{1}_{\{ \rho \in X_t \}} \bigg] \geq \limsup_{t\to\infty} \mathbb{P}(\rho \in X_{t}) > 0,\end{align*}

which means that the process survives strongly.

It remains to show that (6.5) holds. We begin by noting that Condition (6.5) is implied by

(6.17) \begin{equation} \frac{S}{4(2r+1)} > \frac{4}{\widehat{C}_{\lambda, f} \widehat{\lambda}^r}. \end{equation}

Indeed, since $\widehat{\lambda}<1$ , the right-hand side in (6.5) goes to infinity as $r\to \infty$ so that we can ignore the $-1$ on the left-hand side of (6.5). Now, since $C_{\lambda, f}\geq 1$ , we see from the definition of S given in (5.14) with $\epsilon={\frac{5}{16}}$ , that

\begin{equation*} (1-\mathrm{e}^{-\lambda \vartheta fL})(1-\mathrm{e}^{-\gamma})\Big(1+\frac{\lambda f \vartheta}{2}\Big)^{{L(1-2\epsilon)-1}} \Big(\frac{\lambda \vartheta^2}{\lambda \vartheta^2 +1}\Big)^r > 16k(2r +1), \end{equation*}

implies Condition (6.17). Recall that $\epsilon = {\frac{5}{16}}$ , so the latter inequality is equivalent to

\begin{equation*} \log \bigg (\frac{\lambda \vartheta^2}{1+\lambda \vartheta^2} \bigg ) > \frac{1}{r} \bigg (\log \bigg (\frac{16k(2r+1)}{(1-\mathrm{e}^{-\lambda \vartheta fL})(1-\mathrm{e}^{-\gamma})} \bigg )- \bigg ({ \frac{3L}{{8}}}-1 \bigg )\log \bigg (1+\frac{\lambda f \vartheta}{2} \bigg ) \bigg ). \end{equation*}

Taking into account that by the definition of L given in (5.6), $f \kern0.5pt L \rightarrow \infty $ as either $k \rightarrow \infty$ or $f\rightarrow \infty$ , we have

\begin{align*} \frac{1}{r}\log\bigg(\frac{16(2r+1)}{(1-\mathrm{e}^{-\lambda \vartheta fL})(1-\mathrm{e}^{-\gamma})}\bigg) \to 0,\quad \text{as either}\quad k\to \infty\quad \text{or}\quad f\to \infty. \end{align*}

It follows that (6.17) holds if

(6.18) \begin{equation} \log\bigg(\frac{\lambda \vartheta^2}{1+\lambda \vartheta^2}\bigg) > \frac{1}{r} \bigg(\log k-\bigg({\frac{3L}{{8}}}-1\bigg)\log\bigg(1+\frac{\lambda f \vartheta}{2}\bigg)\bigg) \,=\!:\, \Delta(f,k). \end{equation}

As this condition is equivalent to

\begin{align*} \lambda >\frac{\exp \{ \Delta(f,k) \}}{\vartheta^2 ( 1- \exp \{ \Delta(f,k) \}) }, \end{align*}

it suffices to show that $\Delta(f_n,k_n) \rightarrow - \infty$ , for suitably chosen $f_n,k_n$ , to complete the proof.

By Assumption ( B ) and Lemma 14, we have that there exists a sequence $(f_n, k_n)_{n\geq 1}$ such that $f_n + k_n \to \infty $ and

(6.19) \begin{equation} \lim_{n\to \infty} \frac{\log \mathbb{P}(\xi^\vartheta=k_n, \, \mathcal{F} \geq f_n)}{k_n\log\! (1+f_n)}=0.\end{equation}

Note that either

(6.20)

Therefore,

\begin{align*} \log k_n / ({3 L /8} - 1) \log\! (1 + f_n) ) \to 0, \quad \text{as either}\quad k_n\to \infty\quad \text{or}\quad f_n\to \infty. \end{align*}

Thus, we can deduce from the definition of r in (6.16) that $\Delta(f_n,k_n) \to -\infty$ if

\begin{align*} \frac{\big( {3L/8} - 1\big) \log\! (1 + f_n)}{\log \mathbb{P} (\xi^\vartheta = k_n,\, \mathcal{F} \geq f_n )} \to -\infty, \end{align*}

which now follows again by (6.20) directly from (6.19). This, shows that for this choice of $(f_n,k_n)$ we can guarantee that for n large enough $\lambda > \mathrm{e}^{ \Delta(f_n,k_n)}/(\vartheta^2 (1 - \mathrm{e}^{ \Delta(f_n,k_n)}))$ , so that Condition (6.17) holds and the process survives strongly.